MA/CSSE 474 Theory of Computation

Transcription

1 MA/CSSE 474 Theory of Computation CFL Hierarchy CFL Decision Problems Your Questions? Previous class days' material Reading Assignments HW 12 or 13 problems Anything else I have included some slides online that we will not have time to do in class, but may be helpful to you anyway. 1

2 Halting It is possible that a PDA may not halt, never finish reading its input. Let = {a} and consider M = L(M) = {a}: (1, a, ) - (2, a, a) - (3,, ) On any other input except a: M will never halt. M will never finish reading its input unless its input is. Nondeterminism and Decisions 1. There are context-free languages for which no deterministic PDA exists. 2. It is possible that a PDA may not halt, not ever finish reading its input. require time that is exponential in the length of its input. 3. There is no PDA minimization algorithm. It is undecidable whether a PDA is minimal. 2

3 Solutions to the Problem For NDFSMs: Convert to deterministic, or Simulate all paths in parallel. For NDPDAs: No general solution. Formal solutions that usually involve changing the grammar. Such as Chomsky or Greibach Normal form. Practical solutions that: Preserve the structure of the grammar, but Only work on a subset of the CFLs. LL(k), LR(k) (compilers course) What About These Variations? In HW, we see that acceptance by "accepting state only" is equivalent to acceptance by empty stack and accepting state. Equivalent In this sense: Given a language L, there is a PDA that accepts L by accepting state and empty stack iff there is a PDA that accepts L by accepting state only. FSM plus FIFO queue (instead of stack)? FSM plus two stacks? 3

4 Closure Theorems for Context-Free Languages The context-free languages are closed under: Union Concatenation Let G 1 = (V 1, 1, R 1, S 1 ), and G 2 = (V 2, 2, R 2, S 2 ) generate languages L 1 and L 2 Kleene star Reverse Formal details on next slides; we will do them informally Closure Under Union Let G 1 = (V 1, 1, R 1, S 1 ), and G 2 = (V 2, 2, R 2, S 2 ). Assume that G 1 and G 2 have disjoint sets of nonterminals, not including S. Let L = L(G 1 ) L(G 2 ). We can show that L is CF by exhibiting a CFG for it: G = (V 1 V 2 {S}, 1 2, R 1 R 2 {S S 1, S S 2 }, S) 4

5 Closure Under Concatenation Let G 1 = (V 1, 1, R 1, S 1 ), and G 2 = (V 2, 2, R 2, S 2 ). Assume that G 1 and G 2 have disjoint sets of nonterminals, not including S. Let L = L(G 1 )L(G 2 ). We can show that L is CF by exhibiting a CFG for it: G = (V 1 V 2 {S}, 1 2, R 1 R 2 {S S 1 S 2 }, S) Closure Under Kleene Star Let G = (V,, R, S 1 ). Assume that G does not have the nonterminal S. Let L = L(G)*. We can show that L is CF by exhibiting a CFG for it: G = (V 1 {S}, 1, R 1 {S, S SS 1 }, S) 5

6 Closure Under Reverse L R = {w * : w = x R for some x L}. Let G = (V,, R, S) be in Chomsky normal form. Every rule in G is of the form X BC or X a, where X, B, and C are elements of V - and a. X a: L(X) = {a}. {a} R = {a}. X BC: L(X) = L(B)L(C). (L(B)L(C)) R = L(C) R L(B) R. Construct, from G, a new grammar G, such that L(G ) = L R : G = (V G, G, R, S G ), where R is constructed as follows: For every rule in G of the form X BC, add to R the rule X CB. For every rule in G of the form X a, add to R the rule X a. Closure Under Intersection The context-free languages are not closed under intersection: The proof is by counterexample. Let: L 1 = {a n b n c m : n, m 0} L 2 = {a m b n c n : n, m 0} /* equal a s and b s. /* equal b s and c s. Both L 1 and L 2 are context-free, since there exist straightforward context-free grammars for them. But now consider: L = L 1 L 2 = {a n b n c n : n 0} Recall: Closed under union but not closed under intersection implies not closed under complement. And we saw a specific example of a CFL whose complement was not CF. 6

7 The Intersection of a Context-Free Language and a Regular Language is Context-Free L = L(M 1 ), a PDA = (K 1,, 1, 1, s 1, A 1 ). R = L(M 2 ), a deterministic FSM = (K 2,,, s 2, A 2 ). We construct a new PDA, M 3, that accepts L R by simulating the parallel execution of M 1 and M 2. M = (K 1 K 2,, 1,, [s 1, s 2 ], A 1 A 2 ). I use square brackets for ordered pairs of Insert into : states from K 1 K 2, to For each rule ((q 1, a, ), (p 1, )) in 1, distinguish them from and each rule ( q 2, a, p 2 ) in, the tuples that are contains (([q 1, q 2 ] a, ), ([p 1, p 2 ], )). part of the notations for transitions in M 1, For each rule ((q 1,, ), (p 1, ) in 1, M 2, and M. and each state q 2 in K 2, contains (([q 1, q 2 ],, ), ([p 1, q 2 ], )). This works because: we can get away with only one stack. The Difference between a Context-Free Language and a Regular Language is Context-Free Theorem: The difference (L 1 L 2 ) between a context-free language L 1 and a regular language L 2 is context-free. Proof: L 1 L 2 = L 1 L 2. If L 2 is regular then so is L 2. If L 1 is context-free, so is L 1 L 2. 7

8 Deterministic PDAs A PDA M is deterministic iff: M contains no pairs of transitions that compete with each other, and Whenever M is in an accepting configuration it has no available moves. / / M can choose between accepting and taking the -transition, so it is not deterministic. Deterministic CFLs (very quick overview without many details) A language L is deterministic context-free iff L$ can be accepted by some deterministic PDA. Why $? Let L = a* {a n b n : n > 0}. 8

9 An NDPDA for L L = a* {a n b n : n > 0}. A DPDA for L$ L = a* {a n b n : n > 0}. 9

10 DCFL Properties (skip the details). The Deterministic CF Languages are closed under complement. The Deterministic CF Languages are not closed under intersection or union. Nondeterministic CFLs Theorem: There exist CLFs that are not deterministic. Proof: By example. Let L = {a i b j c k, i j or j k}. L is CF. If L is DCF then so is: L = L. = {a i b j c k, i, j, k 0 and i = j = k} {w {a, b, c}* : the letters are out of order}. But then so is: L = L a*b*c*. = {a n b n c n, n 0}. But it isn t. So L is CF but not DCF. This simple fact poses a real problem for the designers of efficient context-free parsers. Solution: design a language that is deterministic. LL(k) or LR(k). 10

11 The CFL Hierarchy Context-Free Languages Over a Single-Letter Alphabet Theorem: Any context-free language over a single-letter alphabet is regular. Proof: Requires Parikh s Theorem, which we are skipping 11

12 Algorithms and Decision Procedures for Context-Free Languages Chapter 14 Decision Procedures for CFLs Membership: Given a language L and a string w, is w in L? Two approaches: If L is context-free, then there exists some context-free grammar G that generates it. Try derivations in G and see whether any of them generates w. Problem (later slide): If L is context-free, then there exists some PDA M that accepts it. Run M on w. Problem (later slide): 12

13 Decision Procedures for CFLs Membership: Given a language L and a string w, is w in L? Two approaches: If L is context-free, then there exists some context-free grammar G that generates it. Try derivations in G and see whether any of them generates w. S S T a Try to derive aaa S S T S T Decision Procedures for CFLs Membership: Given a language L and a string w, is w in L? If L is context-free, then there exists some PDA M that accepts it. Run M on w. Problem: 13

14 Using a Grammar decidecflusinggrammar(l: CFL, w: string) = 1. If given a PDA, build G so that L(G) = L(M). 2. If w = then if S G is nullable then accept, else reject. 3. If w then: 3.1 Construct G in Chomsky normal form such that L(G ) = L(G) { }. 3.2 If G' derives w, it does so in steps. Try all derivations in G' of steps. If one of them derives w, accept. Otherwise reject. How many steps (as a function of w ) in the derivation of w from CNF grammar G'? Using a Grammar decidecflusinggrammar(l: CFL, w: string) = 1. If given a PDA, build G so that L(G) = L(M). 2. If w = then if S G is nullable then accept, else reject. 3. If w then: 3.1 Construct G in Chomsky normal form such that L(G ) = L(G) { }. 3.2 If G' derives w, it does so in 2 w - 1 steps. Try all derivations in G' of 2 w - 1 steps. If one of them derives w, accept. Otherwise reject. Alternative O(n 3 ) algorithm: CKY. a.k.a. CYK. 14

15 Membership Using a PDA Recall CFGtoPDAtopdown, which built: M = ({p, q},, V,, p, {q}), where contains: The start-up transition ((p,, ), (q, S)). For each rule X s 1 s 2 s n. in R, the transition ((q,, X), (q, s 1 s 2 s n )). For each character c, the transition ((q, c, c), (q, )). Can we make this work so there are no -transitions? If every transition consumes an input character then M would have to halt after w steps. Put the grammar into Greibach Normal Form: All rules are of the following form: X a γ, where a and γ (V - )*. We can combine pushing the RHS of the production with matching the first character a. Details on p Greibach Normal Form All rules are of the following form: X aa, where a and A (V - )*. No need to push the a and then immediately pop it. So M = ({p, q},, V,, p, {q}), where contains: 1. The start-up transitions: For each rule S cs 2 s n, the transition: ((p, c, ), (q, s 2 s n )). 2. For each rule X cs 2 s n (where c and s 2 through s n are elements of V - ), the transition: ((q, c, X), (q, s 2 s n )) 15

16 A PDA Without -Transitions Must Halt Consider the execution of M on input w: Each individual path of M must halt within w steps. The total number of paths pursued by M must be less than or equal to P = B w, where B is the maximum number of competing transitions from any state in M. The total number of steps that will be executed by all paths of M is bounded by P w. So all paths must eventually halt. Emptiness Given a context-free language L, is L =? decidecflempty(g: context-free grammar) = 1. Let G = removeunproductive(g). 2. If S is not present in G then return True else return False. 16

17 Finiteness Given a context-free language L, is L infinite? decidecflinfinite(g: context-free grammar) = 1. Lexicographically enumerate all strings in * of length greater than b n and less than or equal to b n+1 + b n. 2. If, for any such string w, decidecfl(l, w) returns True then return True. L is infinite. 3. If, for all such strings w, decidecfl(l, w) returns False then return False. L is not infinite. Why these bounds? Some Undecidable Questions about CFLs Is L = *? Is the complement of L context-free? Is L regular? Is L 1 = L 2? Is L 1 L 2? Is L 1 L 2 =? Is L inherently ambiguous? Is G ambiguous? 17

18 Regular and CF Languages Regular Languages regular exprs. or regular grammars = DFSMs recognize minimize FSMs closed under: concatenation union Kleene star complement intersection pumping theorem D = ND Context-Free Languages context-free grammars = NDPDAs parse try to find unambiguous grammars try to reduce nondeterminism in PDAs find efficient parsers closed under: concatenation union Kleene star intersection w/ reg. langs pumping theorem D ND TURING MACHINE INTRO 18

19 Languages and Machines SD D Context-Free Languages Regular Languages reg exps FSMs cfgs PDAs unrestricted grammars Turing Machines Grammars, SD Languages, and Turing Machines L SD Language Unrestricted Grammar Accepts Turing Machine 19

20 Turing Machines (TMs) We want a new kind of automaton: powerful enough to describe all computable things, unlike FSMs and PDAs. simple enough that we can reason formally about it like FSMs and PDAs, unlike real computers. Goal: Be able to prove things about what can and cannot be computed. Turing Machines At each step, the machine must: choose its next state, write on the current square, and move left or right. 20

21 A Formal Definition A (deterministic) Turing machine M is (K,,,, s, H): K is a finite set of states; is the input alphabet, which does not contain ; is the tape alphabet, which must contain and have as a subset. s K is the initial state; H K is the set of halting states; is the transition function: (K - H) to K {, } non-halting tape state tape direction to move state char char (R or L) Notes on the Definition 1. The input tape is infinite in both directions. 2. is a function, not a relation. So this is a definition for deterministic Turing machines. 3. must be defined for all (state, tape symbol) pairs unless the state is a halting state. 4. Turing machines do not necessarily halt (unlike FSM's and most PDAs). Why? To halt, they must enter a halting state. Otherwise they loop. 5. Turing machines generate output, so they can compute functions. 21

22 An Example M takes as input a string in the language: {a i b j, 0 j i}, and adds b s as required to make the number of b s equal the number of a s. The input to M will look like this: The output should be: The Details (ε) 1 22

23 The Details (ε) 2 The Details (ε) 6 23

24 The Details(aa) a a 1 The Details(aa) a a 2 24

25 The Details(aa) $ a 3 The Details(aa) $ a 3 25

26 The Details(aa) $ a # 4 The Details(aa) $ $ # 3 26

27 The Details(aa) $ $ # 3 The Details(aa) $ $ # 3 27

28 The Details(aa) $ $ # # 4 The Details(aa) $ $ # # 4 28

29 The Details(aa) $ $ # # 4 The Details(aa) $ $ # # 4 29

30 The Details(aa) $ $ # # 5 The Details(aa) a $ # # 5 30

31 The Details(aa) a a # # 5 The Details(aa) a a b # 5 31

32 The Details(aa) a a b b 5 The Details(aa) a a b b 6 32

33 The Details (aabb) a a b b 1 The steps are the same as previous example until we read the b; skip to there The Details (aabb) ε, aa, aabb, a, aab, b $ a b b 3 33

34 The Details (aabb) $ a # b 4 The Details (aabb) $ $ # b 3 34

35 The Details (aabb) $ $ # b 3 The Details (aabb) $ $ # # In state 4, go left until we hit a blank 4 35

36 The Details (aabb) ε, aa, aabb, a, aab, b $ $ # # 4 The Details (aabb) $ $ # # 4 36

37 The Details (aabb) $ $ # # 5 The Details (aabb) $ $ # # 5 Go right, replacing $ with a and # with b, then move left, as in the last part of the previous example. 37

38 The Details(aabb) a a b b 6 The Details (aab) a a b 1 You should try this one. 38

39 The Details (b) b 1 In the first step, move right. Then there is no transition in the diagram. But there is an implied transition to a dead state, i.e. a new halting state that does not accept. Notes on Programming The machine has a strong procedural feel, with one phase coming after another. There are common idioms, like scan left until you find a blank There are two common ways to scan back and forth marking things off. Often there is a final phase to fix up the output. Even a very simple machine is a nuisance to write. 39